copy_from_user in SKAS mode

uml-devel Young Koh 3 participants 6 messages May 17, 2005 – May 19, 2005 User-mode Linux (UML)

Share & tools

Young Koh

author

2005-05-17 23:33 UTC #0 lore raw

Hi,

i'm trying to understand how copy_from_user (and copy_to_user) is working in SKAS mode. i started from copy_from_user_skas, and followed the functions. but i kinda got lost and couldn't get the intuition how data are transferred to the UML kernel, which is a different process. Could anyone explain how copy_from_user works in SKAS mode? i'd really appreciate it. Thank you!

-------------------------------------------------------
This SF.Net email is sponsored by Oracle Space Sweepstakes Want to be the first software developer in space? Enter now for the Oracle Space Sweepstakes!
http://ads.osdn.com/?ad_idt12&alloc_id344&opÌk
_______________________________________________
User-mode-linux-devel mailing list [email protected]
https://lists.sourceforge.net/lists/listinfo/user-mode-linux-devel

Jeff Dike

2005-05-18 01:23 UTC #1 lore raw

Quoted 6 lines · Young Koh

i'm trying to understand how copy_from_user (and copy_to_user) is working in SKAS mode. i started from copy_from_user_skas, and followed the functions. but i kinda got lost and couldn't get the intuition how data are transferred to the UML kernel, which is a different process. Could anyone explain how copy_from_user works in SKAS mode? i'd really appreciate it. Thank you!

Since UML has no direct access to the process userspace (i.e. process virtual addresses are meaningless in the kernel), it has to translate them to physical addresses. So, each page containing a piece of the incoming buffer is translated into a physical page, and the data copied from there.

Jeff

-------------------------------------------------------
This SF.Net email is sponsored by Oracle Space Sweepstakes Want to be the first software developer in space? Enter now for the Oracle Space Sweepstakes!
http://ads.osdn.com/?ad_id=7412&alloc_id=16344&op=click
_______________________________________________
User-mode-linux-devel mailing list [email protected]
https://lists.sourceforge.net/lists/listinfo/user-mode-linux-devel

Young Koh

author

2005-05-18 13:47 UTC #2 lore raw

Quoted 5 lines

Since UML has no direct access to the process userspace (i.e. process virtual addresses are meaningless in the kernel), it has to translate them to physical addresses. So, each page containing a piece of the incoming buffer is translated into a physical page, and the data copied from there.

you mean "physical" to the UML kernel? Because the UML kernel cannot know about the real physical pages in the host kernel, do you mean the UML kernel finds out which its (virtual) address maps to the same real physical page that the user process's address maps to? but the UML kernel and the user process would have different mappings and cannot know about each other (if so, its a protection violation between processes) Could you explain a bit more? Thanks a lot!!!

-Young

Quoted 3 lines

Jeff

Jeff Dike

2005-05-18 15:13 UTC #3 lore raw

Quoted 1 line · Young Koh

you mean "physical" to the UML kernel?

Yes. To the host, it's just normal process virtual memory, but to UML, that's its physical memory.

Quoted 7 lines

Because the UML kernel cannot
know about the real physical pages in the host kernel, do you mean the UML kernel finds out which its (virtual) address maps to the same real physical page that the user process's address maps to? but the UML kernel and the user process would have different mappings and cannot know about each other (if so, its a protection violation between processes) Could you explain a bit more? Thanks a lot!!!

It's exactly the same as the host. There is physical memory mapped into its address space, and pages from that area are allocated and mapped into process address spaces as needed. In skas mode, UML has no direct access to the process address spaces (like the x86 4G/4G split), so it has to translate the process virtual address into a UML physical address, to which it does have access, and can copy the data.

Jeff

-------------------------------------------------------
This SF.Net email is sponsored by Oracle Space Sweepstakes Want to be the first software developer in space? Enter now for the Oracle Space Sweepstakes!
http://ads.osdn.com/?ad_id=7412&alloc_id=16344&op=click
_______________________________________________
User-mode-linux-devel mailing list [email protected]
https://lists.sourceforge.net/lists/listinfo/user-mode-linux-devel

Young Koh

author

2005-05-18 22:34 UTC #4 lore raw

Ok, let me have an example.

suppose a user argument has virtual address 0xa0, corresponding UML physical address 0xb0, and real physical address 0x10. so, when the user process tries to access 0xa0 for the first time, the UML kernel should let host kernel know there will be 0xa0 -> 0x10 mapping. how does the UML kernel make another process's address map to the same physical page it has? and how does the UML kernel keep track of 0xa0 -> 0xb0 mapping? maybe the same page table mechanism as host Linux? (i guess so, though)

Thanks a lot!

Quoted 14 lines · Jeff Dike

Quoted 1 line · Young Koh
you mean "physical" to the UML kernel?
Yes. To the host, it's just normal process virtual memory, but to UML, that's its physical memory.
Quoted 7 lines
Because the UML kernel cannot
know about the real physical pages in the host kernel, do you mean the UML kernel finds out which its (virtual) address maps to the same real physical page that the user process's address maps to? but the UML kernel and the user process would have different mappings and cannot know about each other (if so, its a protection violation between processes) Could you explain a bit more? Thanks a lot!!!

Quoted 9 lines

It's exactly the same as the host. There is physical memory mapped into its address space, and pages from that area are allocated and mapped into process address spaces as needed. In skas mode, UML has no direct access to the process address spaces (like the x86 4G/4G split), so it has to translate the process virtual address into a UML physical address, to which it does have access, and can copy the data.
Jeff

Blaisorblade

2005-05-19 13:42 UTC #5 lore raw

Quoted 6 lines · Young Koh

Ok, let me have an example.
suppose a user argument has virtual address 0xa0, corresponding UML physical address 0xb0, and real physical address 0x10. so, when the user process tries to access 0xa0 for the first time, the UML kernel should let host kernel know there will be 0xa0 -> 0x10 mapping.

Yes, and we only need to know the 0xa0 -> 0xb0 mapping... We currently use a 2/3-level page table mechanism to do this lookup, to answer the latter question, even if we could choice any lookup mechanism, even a hash table if there were the need.

Also note that these numbers are unrealistic since we map one page (4k) at a time.

Quoted 3 lines

how
does the UML kernel make another process's address map to the same physical page it has?

UML physical address are actually the offsets inside the memory backing file. Then the guest has simply to execute an mmap from this file with the "Uml physical address" as the offset; in TT mode it's done directly, in SKAS3 mode this is done by writing a MM_MMAP request into host's /proc/mm, in SKAS0 this is executed by the stub calling mmap() directly.

Quoted 3 lines

and how does the UML kernel keep track of 0xa0
-> 0xb0 mapping? maybe the same page table mechanism as host Linux? (i guess so, though)

Signature · 17 lines

Paolo Giarrusso, aka Blaisorblade
Skype user "PaoloGiarrusso"
Linux registered user n. 292729
http://www.user-mode-linux.org/~blaisorblade




-------------------------------------------------------
This SF.Net email is sponsored by Oracle Space Sweepstakes
Want to be the first software developer in space?
Enter now for the Oracle Space Sweepstakes!
http://ads.osdn.com/?ad_id=7412&alloc_id=16344&op=click
_______________________________________________
User-mode-linux-devel mailing list
[email protected]
https://lists.sourceforge.net/lists/listinfo/user-mode-linux-devel